Question: Let $P$ be a point inside triangle $ABC$.  Let $G_1$, $G_2$, and $G_3$ be the centroids of triangles $PBC$, $PCA$, and $PAB$, respectively.  If the area of triangle $ABC$ is 18, then find the area of triangle $G_1 G_2 G_3$.

[asy]
import geometry;

unitsize(2 cm);

pair A, B, C, P;
pair[] G;

A = (1,3);
B = (0,0);
C = (4,0);
P = (2,1);
G[1] = (P + B + C)/3;
G[2] = (P + C + A)/3;
G[3] = (P + A + B)/3;

draw(A--B--C--cycle);
draw(A--P);
draw(B--P);
draw(C--P);
draw(G[1]--G[2]--G[3]--cycle);

label("$A$", A, dir(90));
label("$B$", B, SW);
label("$C$", C, SE);
dot("$G_1$", G[1], S);
dot("$G_2$", G[2], SE);
dot("$G_3$", G[3], NW);
label("$P$", P, S);
[/asy]
Answer: Let $M_1$, $M_2$, and $M_3$ be the midpoints of $AP$, $BP$, and $CP$, respectively.  Then as a midline in triangle $PBC$, $M_2 M_3$ is parallel to $BC$, and half the length of $BC$.

[asy]
import geometry;

unitsize(2 cm);

pair A, B, C, P;
pair[] G, M;

A = (1,3);
B = (0,0);
C = (4,0);
P = (2,1);
G[1] = (P + B + C)/3;
G[2] = (P + C + A)/3;
G[3] = (P + A + B)/3;
M[1] = (P + A)/2;
M[2] = (P + B)/2;
M[3] = (P + C)/2;

draw(A--B--C--cycle);
draw(A--P);
draw(B--P);
draw(C--P);
draw(A--M[2]);
draw(A--M[3]);
draw(G[2]--G[3]);
draw(M[2]--M[3]);

label("$A$", A, dir(90));
label("$B$", B, SW);
label("$C$", C, SE);
dot("$G_2$", G[2], NE);
dot("$G_3$", G[3], W);
dot("$M_2$", M[2], S);
dot("$M_3$", M[3], S);
label("$P$", P, S);
[/asy]

Since $G_3$ is the centroid of triangle $PAB$, $G_3$ divides median $AM_2$ in the ratio $2:1$.  Similarly, $G_2$ divides median $AM_3$ in the ratio $2:1$.  Therefore, triangles $AG_3 G_2$ and $AM_2 M_3$ are similar.  Also, $G_2 G_3$ is parallel to $M_2 M_3$, and $G_2 G_3$ is two-thirds the length of $M_2 M_3$.

Therefore, $G_2 G_3$ is parallel to $BC$, and $G_2 G_3$ is one-third the length of $BC$.  Likewise, $G_1 G_2$ is parallel to $AB$, and $G_1 G_2$ is one-third the length of $AB$.  Hence, triangle $G_1 G_2 G_3$ is similar to triangle $ABC$, with ratio of similarity 1/3.  The area of triangle $ABC$ is 18, so the area of triangle $G_1 G_2 G_3$ is $18 \cdot (1/3)^2 = \boxed{2}$.